Solve left(a+bright)^2=left(a+bright)left(a+bright)=

$${ \left(a+b \right) }^{ 2 } = \left( a+b \right) \left( a+b \right) =$$

Solve for a (complex solution)

$$a\in \mathrm{C}$$

Solution Steps

Multiply a+b and a+b to get \left(a+b\right)^{2}.

$$\left(a+b\right)^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=a^{2}+2ab+b^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+2ab+b^{2}-a^{2}=2ab+b^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$2ab+b^{2}=2ab+b^{2}$$

Subtract 2ab from both sides.

$$2ab+b^{2}-2ab=b^{2}$$

Combine 2ab and -2ab to get 0.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{C}$$

Solve for b (complex solution)

$$b\in \mathrm{C}$$

Solution Steps

Multiply a+b and a+b to get \left(a+b\right)^{2}.

$$\left(a+b\right)^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=a^{2}+2ab+b^{2}$$

Subtract 2ab from both sides.

$$a^{2}+2ab+b^{2}-2ab=a^{2}+b^{2}$$

Combine 2ab and -2ab to get 0.

$$a^{2}+b^{2}=a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$a^{2}+b^{2}-b^{2}=a^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{C}$$

Solve for a

$$a\in \mathrm{R}$$

Solution Steps

Multiply a+b and a+b to get \left(a+b\right)^{2}.

$$\left(a+b\right)^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=a^{2}+2ab+b^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+2ab+b^{2}-a^{2}=2ab+b^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$2ab+b^{2}=2ab+b^{2}$$

Subtract 2ab from both sides.

$$2ab+b^{2}-2ab=b^{2}$$

Combine 2ab and -2ab to get 0.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for b

$$b\in \mathrm{R}$$

Solution Steps

Multiply a+b and a+b to get \left(a+b\right)^{2}.

$$\left(a+b\right)^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=\left(a+b\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}+2ab+b^{2}=a^{2}+2ab+b^{2}$$

Subtract 2ab from both sides.

$$a^{2}+2ab+b^{2}-2ab=a^{2}+b^{2}$$

Combine 2ab and -2ab to get 0.

$$a^{2}+b^{2}=a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$a^{2}+b^{2}-b^{2}=a^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{R}$$